3+1 formalism and bases of numerical relativity - LUTh ...

120 Asymptotic flatness and global quantities
Example : A simple prototype of a stationary spacetime is of course the Schwarzschild spacetime.
Let us compute its Komar mass by means of the above formula and the foliation
(Σt)t∈R defined by the standard Schwarzschild coordinates (7.16). For this foliation,
Kij = 0, which reduces Eq. (7.91) to the flux of the lapse’s gradient across St. Taking
advantage of the spherical symmetry, we choose St to be a surface r = const. Then
y a = (θ,ϕ). The unit normal s is read from the line element (7.16); its components with
respect to the Schwarzschild coordinates (r,θ,ϕ) are
s i =
1 − 2m
1/2
,0,0
r
. (7.92)
N and √ q are also read on the line element (7.16): N = (1 − 2m/r) 1/2 and √ q = r2 sinθ,
so that Eq. (7.91) results in
MK = 1
1 −
4π r=const
2m
1/2
∂
1 −
r ∂r
2m
1/2
r
r
2 sin θdθdϕ. (7.93)
All the terms containing r simplify and we get
MK = m. (7.94)
On this particular example, we have verified that the value of MK does not depend upon
the choice of St.
Let us now turn to the volume expression (7.87) of the Komar mass. By using the 3+1
decomposition (4.10) and (4.12) of respectively T and T, we get
T(n,k) − 1
1
T n · k = −〈p,k〉 − E〈n,k〉 − (S − E)n · k
2 2
= −〈p,β〉 + EN + 1 1
(S − E)N = N(E + S) − 〈p,β〉. (7.95)
2 2
Hence formula (7.87) becomes
MK =
Σt
[N(E + S) − 2〈p,β〉] √ γ d 3 x + M H K
, (7.96)
with the Komar mass of the hole given by an expression identical to Eq. (7.91), except for St
replaced by Ht [notice the double change of sign: first in Eq. (7.83) and secondly in Eq. (7.76),
so that at the end we have an expression identical to Eq. (7.91)]:
M H K = 1
4π
i
s DiN − Kijs i β j √ 2
q d y . (7.97)
Ht
It is easy to take the Newtonian limit Eq. (7.96), by making N → 1, E → ρ, S ≪ E
[Eq. (5.25)], β → 0, γ → f and M H K = 0. We get
MK = ρ f d 3 x. (7.98)
Σt